#$&* course Mth 277 1210pm 12/8/2011 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: The vector field is conservative if, and only if, its curl is zero. The curl of the field M i + N j + P k is (P_y - N_z) i + (M_z - P_x) j + (N_x - M_y) k. For the current function there is no z dependence and P = 0, so the expression becomes just (N_x - M_y) k. M_y = 1 and N_x = 2, so the field is not conservative. If we did try to find the scalar potential function, we would try to find f(x, y) such that del f = F. If this was so then we would have f_x = M and f_y = N. If f_x = (y - x^2), then integrating with respect to x we conclude that f = y x - x^3 / 3 + g(y), where g(y) is any function of y. If f_y = (2 x + y^2), then integrating with respect to y we conclude that f = 2 x y + y^3 / 3 + h(x), where h(x) is any function of x. We could let g(y) = y^3 / 3, and h(x) = - x^3 / 3. However the first integration yields y x, and the second yields 2 x y, both functions of x as well as y. That is, any f function such that del f = F must contain the term x y as well as the term 2 x y. We can't accomplish this by adjusting either g(y) or h(x), and we conclude that no scalar potential function f exists for this F. Had the original function been F(x, y) = (2 y - x^2) i + (2 x + y^2) j then : Our integrals would have been 2 y x - x^3 / 3 + g(y) and 2 x y + y^3 / 3 + h(x). Letting g(y) = y^3 / 3 and h(x) = -x^3 / 3, our function would be the same using either the x or the y integral; either way we would get 2 x y - x^3 / 3 + y^3 / 3. M_y would have been 2 and N_x would have been 2, so that our curl would have been (N_x - M_y) k = 0. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ********************************************* Question: Show that F = e^-y i - xe^-y j is conservative and find a scalar potential f for F. Evaluate the line integral Int[ F dot dR, C], where C is any smooth path connecting (0,0) and (1,1). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: M = e^-y N = -xe^-y dM/dy = -e^-y dN/dx = -e^-y Therefore the vector field is conservative f(x,y) = Int[f_x(x,y)dx] = Int[e^-y dx] = xe^-y + C1 f(x,y) = Int[f_y(x,y)dy] = Int[-xe^-y dy] = xe^-y + C2 f(x,y) = xe^-y Int[F dot dR, C] R = ti + tj, where 0 =< t =< 1 dR = i + j Int[(e^-y i - xe^-y j) dot (i + j)dt = Int[(e^-t - te^-t) dt from 0 to 1 u = -t, dv = e^-tdt du = dt, v = -e^-t = -e^-t + te^-t + Int[e^-tdt from 0 to 1 = -e^-t + te^-t - te^-t from 0 to 1 = -e^-t from 0 to 1 = -e^-1 + e^0 =appr 0.632 = -te^-t confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The curl of this function, which is of the form M i + N j + P k with M = e^-y, N = - x e^(-y) and P = 0, is easily found to be (P_y - N_z) i + (M_z - P_x) j + (N_x - M_y) k = (N_x - M_y) k = ( -e^-y - (-e^(-y)) ) k = 0. So the function is conservative and a scalar potential function exists. We therefore integrate: Integrating e^-y with respect to x we get x e^-y + g(y). Integrating - x e^(-y) with respect to y we get x e^-y + h(x). The general form of a scalar potential for our given vector field F is therefore f(x, y) = x e^-y + g(y) + h(x), where g(y) and h(x) are arbitrary functions of y and x, respectively. The simplest such function is obtained by letting g(y) and h(x) both be zero, giving us f(x, y) = x e^-y We can easily verify that del f = F. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ********************************************* Question: Consider the integral Int[ 2x^2y dx + x^3 dy, C]. Evaluate it on each of the following curves C: The semicircle x = sqrt(1-y) from -1 <= y <= 1. The line segment from (0,-1) to (1,1) and then the line segment from (1,1) to (0,1). The square with vertices (1,1), (-1,1), (-1,-1), (1,-1) traversed once clockwise. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First check to see F is conservative M = 2x^2*y N = x^3 My = 2x^2 Nx = 3x^2 NOT conservative, so now we have to do each on individually Int[(2x^2)*ydx + x^3dy y(t) = t from -1 to 1 x(t) = sqrt(1-t) dy = dt dx = [-1/sqrt(1-t)]dt Int[(2x^2)*ydx + x^3dy = Int{2t(1-t) - [1/sqrt(1-t)]dt = (2t^2)/2 - (2t^3)/3 + sqrt(1-t) + (2/5)(1-t)^(5/2) from 1 to -1 = [1 - 2/3 + 0 + 0] - [1 + 2/3 + sqrt(2) + (2/5)sqrt(32)] = -1 + sqrt(2) + (4/5)sqrt(8) =app -4.68 units, which is only negative because we are going from 1 to -1. Int[(2x^2)*ydx + x^3dy y - y1 = m(x - x1) y = 2x - 1 y(t) = 2t - 1 x(t) = t dy = 2dt dx = dt The line segment from (0,-1) to (1,1) Int[(2x^2)*ydx + x^3dy = 2t^2(2t-1)dt + 2t^3dt = 4t^3 - 2t^2dt + 2t^3dt =t^4 - (2/3)t^3 + (t^4)/2 from 1 to 0 = 1-(2/3)+(1/2) - 0 = 5/6 and then the line segment from (1,1) to (0,1) Int[(2x^2)*ydx + x^3dy y=1 x = -t + 1 dy = dt dx=-dt Int[(2x^2)*ydx + x^3dy = Int[(2(-t+1)^2)*-dt + (-t+1)^3dt from 0 to 1 =(2/3)(-t+1)^3 - (1/4)(-t+1)^4 from 0 to 1 = 0 - [2/3 -1/4] = -5/12 Therefore 5/6 - 5/12 = 5/12 units Finally the last on The square with vertices (1,1), (-1,1), (-1,-1), (1,-1) traversed once clockwise. To make this easier and doing 4 seperate intergrations, we can just multiply the Integration by four and since this is a nonconservative force it can't equal zero. so therefore the intergral from 1,1 to -1,1 times four is the amount of work done. For the last problem the line segment from (1,1) to (0,1) is -5/12. Multiply that by 2 and then that by 8, we have our answer, which is equal to -3.33 or -10/3 and the negative is just simply because we are going in a counterclockwise direction confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ********************************************* Question: Verify that the integral Int[(xy cos (xy) + sin (xy))dx + (x^2 cos(xy)) dy, C] is independent of path and evaluate it where C is any path from (0, pi/18) to (1, pi/6). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Int[(xy cos (xy) + sin (xy))dx + (x^2 cos(xy)) dy, C] M = (xy cos (xy) + sin (xy)) N = (x^2 cos(xy)) dM/dy = -x^2ysin(xy) + xcos(xy) + cos(xy) dN/dx = -x^2ysin(xy) + 2xcos(xy) Therefore the vector field is conservative Int[xycos(xy) + sin(xy)dx u=xy, dv = cos(xy)dx du = ydx, v = sin(xy)/y xsin(xy) - (2/y)cos(xy) + g(y) Int[ x^2cos(xy) = xsin(xy) + h(y) potential function: xsin(xy) Int[ xsin(xy) from (0, pi/18) to (1, pi/6) 1sin(1*pi/6) - 0 = 1/2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: This integral is of the form F dot ds for F = (x y cos(x y) + sin(x y) ) i + x^2 cos(x y) j, of the form M i + N j. We easily verify that curl F = (N_x - M_y) k = ( (2 x cos(xy) - x^2 y sin(x y)) - (x cos(x y) - x^2 y sin(x y) + x cos(x y) ) ) k = 0. We conclude that a scalar potential function exists, and we integrate to find it: Integrating x y cos(x y) + sin(x y) with respect to x we get cos( x y ) / y + x sin(x y) - cos(x y) / y = x sin(x y), to which we can add the arbitrary integration 'constant' g(y). Integrating x^2 cos(x y) with respect to y we get x^2 ( sin(x y) / x) = x sin(x y) + h(x). Our general scalar potential is therefore f(x, y) = x sin(x y) + g(y) + h(x), for arbitrary g(y) and h(x). The simplest scalar potential function is just f(x, y) = x sin(x y), and we will use this to evaluate our integral. For conservative field F our fundamental theorem says that integral ( F dot ds , between points (x0, y0) and (xf, yf) ) = f(xf, yf) - f(x0, y0) Our current integral is conservative. Its path originates at (0, pi/18) and terminates at (1, pi/6), so Int[(xy cos (xy) + sin (xy))dx + (x^2 cos(xy)) dy, C] = f(1, pi/6) - f(0, pi/18) = 1 sin(1 * pi / 6) - 0 sin(0 * pi/18) = 1/2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK" Self-critique (if necessary): ------------------------------------------------ Self-critique rating:
#$&* course Mth 277 1210pm 12/8/2011 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: The vector field is conservative if, and only if, its curl is zero. The curl of the field M i + N j + P k is (P_y - N_z) i + (M_z - P_x) j + (N_x - M_y) k. For the current function there is no z dependence and P = 0, so the expression becomes just (N_x - M_y) k. M_y = 1 and N_x = 2, so the field is not conservative. If we did try to find the scalar potential function, we would try to find f(x, y) such that del f = F. If this was so then we would have f_x = M and f_y = N. If f_x = (y - x^2), then integrating with respect to x we conclude that f = y x - x^3 / 3 + g(y), where g(y) is any function of y. If f_y = (2 x + y^2), then integrating with respect to y we conclude that f = 2 x y + y^3 / 3 + h(x), where h(x) is any function of x. We could let g(y) = y^3 / 3, and h(x) = - x^3 / 3. However the first integration yields y x, and the second yields 2 x y, both functions of x as well as y. That is, any f function such that del f = F must contain the term x y as well as the term 2 x y. We can't accomplish this by adjusting either g(y) or h(x), and we conclude that no scalar potential function f exists for this F. Had the original function been F(x, y) = (2 y - x^2) i + (2 x + y^2) j then : Our integrals would have been 2 y x - x^3 / 3 + g(y) and 2 x y + y^3 / 3 + h(x). Letting g(y) = y^3 / 3 and h(x) = -x^3 / 3, our function would be the same using either the x or the y integral; either way we would get 2 x y - x^3 / 3 + y^3 / 3. M_y would have been 2 and N_x would have been 2, so that our curl would have been (N_x - M_y) k = 0. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ********************************************* Question: Show that F = e^-y i - xe^-y j is conservative and find a scalar potential f for F. Evaluate the line integral Int[ F dot dR, C], where C is any smooth path connecting (0,0) and (1,1). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: M = e^-y N = -xe^-y dM/dy = -e^-y dN/dx = -e^-y Therefore the vector field is conservative f(x,y) = Int[f_x(x,y)dx] = Int[e^-y dx] = xe^-y + C1 f(x,y) = Int[f_y(x,y)dy] = Int[-xe^-y dy] = xe^-y + C2 f(x,y) = xe^-y Int[F dot dR, C] R = ti + tj, where 0 =< t =< 1 dR = i + j Int[(e^-y i - xe^-y j) dot (i + j)dt = Int[(e^-t - te^-t) dt from 0 to 1 u = -t, dv = e^-tdt du = dt, v = -e^-t = -e^-t + te^-t + Int[e^-tdt from 0 to 1 = -e^-t + te^-t - te^-t from 0 to 1 = -e^-t from 0 to 1 = -e^-1 + e^0 =appr 0.632 = -te^-t confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The curl of this function, which is of the form M i + N j + P k with M = e^-y, N = - x e^(-y) and P = 0, is easily found to be (P_y - N_z) i + (M_z - P_x) j + (N_x - M_y) k = (N_x - M_y) k = ( -e^-y - (-e^(-y)) ) k = 0. So the function is conservative and a scalar potential function exists. We therefore integrate: Integrating e^-y with respect to x we get x e^-y + g(y). Integrating - x e^(-y) with respect to y we get x e^-y + h(x). The general form of a scalar potential for our given vector field F is therefore f(x, y) = x e^-y + g(y) + h(x), where g(y) and h(x) are arbitrary functions of y and x, respectively. The simplest such function is obtained by letting g(y) and h(x) both be zero, giving us f(x, y) = x e^-y We can easily verify that del f = F. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ********************************************* Question: Consider the integral Int[ 2x^2y dx + x^3 dy, C]. Evaluate it on each of the following curves C: The semicircle x = sqrt(1-y) from -1 <= y <= 1. The line segment from (0,-1) to (1,1) and then the line segment from (1,1) to (0,1). The square with vertices (1,1), (-1,1), (-1,-1), (1,-1) traversed once clockwise. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First check to see F is conservative M = 2x^2*y N = x^3 My = 2x^2 Nx = 3x^2 NOT conservative, so now we have to do each on individually Int[(2x^2)*ydx + x^3dy y(t) = t from -1 to 1 x(t) = sqrt(1-t) dy = dt dx = [-1/sqrt(1-t)]dt Int[(2x^2)*ydx + x^3dy = Int{2t(1-t) - [1/sqrt(1-t)]dt = (2t^2)/2 - (2t^3)/3 + sqrt(1-t) + (2/5)(1-t)^(5/2) from 1 to -1 = [1 - 2/3 + 0 + 0] - [1 + 2/3 + sqrt(2) + (2/5)sqrt(32)] = -1 + sqrt(2) + (4/5)sqrt(8) =app -4.68 units, which is only negative because we are going from 1 to -1. Int[(2x^2)*ydx + x^3dy y - y1 = m(x - x1) y = 2x - 1 y(t) = 2t - 1 x(t) = t dy = 2dt dx = dt The line segment from (0,-1) to (1,1) Int[(2x^2)*ydx + x^3dy = 2t^2(2t-1)dt + 2t^3dt = 4t^3 - 2t^2dt + 2t^3dt =t^4 - (2/3)t^3 + (t^4)/2 from 1 to 0 = 1-(2/3)+(1/2) - 0 = 5/6 and then the line segment from (1,1) to (0,1) Int[(2x^2)*ydx + x^3dy y=1 x = -t + 1 dy = dt dx=-dt Int[(2x^2)*ydx + x^3dy = Int[(2(-t+1)^2)*-dt + (-t+1)^3dt from 0 to 1 =(2/3)(-t+1)^3 - (1/4)(-t+1)^4 from 0 to 1 = 0 - [2/3 -1/4] = -5/12 Therefore 5/6 - 5/12 = 5/12 units Finally the last on The square with vertices (1,1), (-1,1), (-1,-1), (1,-1) traversed once clockwise. To make this easier and doing 4 seperate intergrations, we can just multiply the Integration by four and since this is a nonconservative force it can't equal zero. so therefore the intergral from 1,1 to -1,1 times four is the amount of work done. For the last problem the line segment from (1,1) to (0,1) is -5/12. Multiply that by 2 and then that by 8, we have our answer, which is equal to -3.33 or -10/3 and the negative is just simply because we are going in a counterclockwise direction confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ********************************************* Question: Verify that the integral Int[(xy cos (xy) + sin (xy))dx + (x^2 cos(xy)) dy, C] is independent of path and evaluate it where C is any path from (0, pi/18) to (1, pi/6). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Int[(xy cos (xy) + sin (xy))dx + (x^2 cos(xy)) dy, C] M = (xy cos (xy) + sin (xy)) N = (x^2 cos(xy)) dM/dy = -x^2ysin(xy) + xcos(xy) + cos(xy) dN/dx = -x^2ysin(xy) + 2xcos(xy) Therefore the vector field is conservative Int[xycos(xy) + sin(xy)dx u=xy, dv = cos(xy)dx du = ydx, v = sin(xy)/y xsin(xy) - (2/y)cos(xy) + g(y) Int[ x^2cos(xy) = xsin(xy) + h(y) potential function: xsin(xy) Int[ xsin(xy) from (0, pi/18) to (1, pi/6) 1sin(1*pi/6) - 0 = 1/2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: This integral is of the form F dot ds for F = (x y cos(x y) + sin(x y) ) i + x^2 cos(x y) j, of the form M i + N j. We easily verify that curl F = (N_x - M_y) k = ( (2 x cos(xy) - x^2 y sin(x y)) - (x cos(x y) - x^2 y sin(x y) + x cos(x y) ) ) k = 0. We conclude that a scalar potential function exists, and we integrate to find it: Integrating x y cos(x y) + sin(x y) with respect to x we get cos( x y ) / y + x sin(x y) - cos(x y) / y = x sin(x y), to which we can add the arbitrary integration 'constant' g(y). Integrating x^2 cos(x y) with respect to y we get x^2 ( sin(x y) / x) = x sin(x y) + h(x). Our general scalar potential is therefore f(x, y) = x sin(x y) + g(y) + h(x), for arbitrary g(y) and h(x). The simplest scalar potential function is just f(x, y) = x sin(x y), and we will use this to evaluate our integral. For conservative field F our fundamental theorem says that integral ( F dot ds , between points (x0, y0) and (xf, yf) ) = f(xf, yf) - f(x0, y0) Our current integral is conservative. Its path originates at (0, pi/18) and terminates at (1, pi/6), so Int[(xy cos (xy) + sin (xy))dx + (x^2 cos(xy)) dy, C] = f(1, pi/6) - f(0, pi/18) = 1 sin(1 * pi / 6) - 0 sin(0 * pi/18) = 1/2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!
#$&* course Mth 277 1210pm 12/8/2011 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: The vector field is conservative if, and only if, its curl is zero. The curl of the field M i + N j + P k is (P_y - N_z) i + (M_z - P_x) j + (N_x - M_y) k. For the current function there is no z dependence and P = 0, so the expression becomes just (N_x - M_y) k. M_y = 1 and N_x = 2, so the field is not conservative. If we did try to find the scalar potential function, we would try to find f(x, y) such that del f = F. If this was so then we would have f_x = M and f_y = N. If f_x = (y - x^2), then integrating with respect to x we conclude that f = y x - x^3 / 3 + g(y), where g(y) is any function of y. If f_y = (2 x + y^2), then integrating with respect to y we conclude that f = 2 x y + y^3 / 3 + h(x), where h(x) is any function of x. We could let g(y) = y^3 / 3, and h(x) = - x^3 / 3. However the first integration yields y x, and the second yields 2 x y, both functions of x as well as y. That is, any f function such that del f = F must contain the term x y as well as the term 2 x y. We can't accomplish this by adjusting either g(y) or h(x), and we conclude that no scalar potential function f exists for this F. Had the original function been F(x, y) = (2 y - x^2) i + (2 x + y^2) j then : Our integrals would have been 2 y x - x^3 / 3 + g(y) and 2 x y + y^3 / 3 + h(x). Letting g(y) = y^3 / 3 and h(x) = -x^3 / 3, our function would be the same using either the x or the y integral; either way we would get 2 x y - x^3 / 3 + y^3 / 3. M_y would have been 2 and N_x would have been 2, so that our curl would have been (N_x - M_y) k = 0. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ********************************************* Question: Show that F = e^-y i - xe^-y j is conservative and find a scalar potential f for F. Evaluate the line integral Int[ F dot dR, C], where C is any smooth path connecting (0,0) and (1,1). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: M = e^-y N = -xe^-y dM/dy = -e^-y dN/dx = -e^-y Therefore the vector field is conservative f(x,y) = Int[f_x(x,y)dx] = Int[e^-y dx] = xe^-y + C1 f(x,y) = Int[f_y(x,y)dy] = Int[-xe^-y dy] = xe^-y + C2 f(x,y) = xe^-y Int[F dot dR, C] R = ti + tj, where 0 =< t =< 1 dR = i + j Int[(e^-y i - xe^-y j) dot (i + j)dt = Int[(e^-t - te^-t) dt from 0 to 1 u = -t, dv = e^-tdt du = dt, v = -e^-t = -e^-t + te^-t + Int[e^-tdt from 0 to 1 = -e^-t + te^-t - te^-t from 0 to 1 = -e^-t from 0 to 1 = -e^-1 + e^0 =appr 0.632 = -te^-t confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The curl of this function, which is of the form M i + N j + P k with M = e^-y, N = - x e^(-y) and P = 0, is easily found to be (P_y - N_z) i + (M_z - P_x) j + (N_x - M_y) k = (N_x - M_y) k = ( -e^-y - (-e^(-y)) ) k = 0. So the function is conservative and a scalar potential function exists. We therefore integrate: Integrating e^-y with respect to x we get x e^-y + g(y). Integrating - x e^(-y) with respect to y we get x e^-y + h(x). The general form of a scalar potential for our given vector field F is therefore f(x, y) = x e^-y + g(y) + h(x), where g(y) and h(x) are arbitrary functions of y and x, respectively. The simplest such function is obtained by letting g(y) and h(x) both be zero, giving us f(x, y) = x e^-y We can easily verify that del f = F. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ********************************************* Question: Consider the integral Int[ 2x^2y dx + x^3 dy, C]. Evaluate it on each of the following curves C: The semicircle x = sqrt(1-y) from -1 <= y <= 1. The line segment from (0,-1) to (1,1) and then the line segment from (1,1) to (0,1). The square with vertices (1,1), (-1,1), (-1,-1), (1,-1) traversed once clockwise. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First check to see F is conservative M = 2x^2*y N = x^3 My = 2x^2 Nx = 3x^2 NOT conservative, so now we have to do each on individually Int[(2x^2)*ydx + x^3dy y(t) = t from -1 to 1 x(t) = sqrt(1-t) dy = dt dx = [-1/sqrt(1-t)]dt Int[(2x^2)*ydx + x^3dy = Int{2t(1-t) - [1/sqrt(1-t)]dt = (2t^2)/2 - (2t^3)/3 + sqrt(1-t) + (2/5)(1-t)^(5/2) from 1 to -1 = [1 - 2/3 + 0 + 0] - [1 + 2/3 + sqrt(2) + (2/5)sqrt(32)] = -1 + sqrt(2) + (4/5)sqrt(8) =app -4.68 units, which is only negative because we are going from 1 to -1. Int[(2x^2)*ydx + x^3dy y - y1 = m(x - x1) y = 2x - 1 y(t) = 2t - 1 x(t) = t dy = 2dt dx = dt The line segment from (0,-1) to (1,1) Int[(2x^2)*ydx + x^3dy = 2t^2(2t-1)dt + 2t^3dt = 4t^3 - 2t^2dt + 2t^3dt =t^4 - (2/3)t^3 + (t^4)/2 from 1 to 0 = 1-(2/3)+(1/2) - 0 = 5/6 and then the line segment from (1,1) to (0,1) Int[(2x^2)*ydx + x^3dy y=1 x = -t + 1 dy = dt dx=-dt Int[(2x^2)*ydx + x^3dy = Int[(2(-t+1)^2)*-dt + (-t+1)^3dt from 0 to 1 =(2/3)(-t+1)^3 - (1/4)(-t+1)^4 from 0 to 1 = 0 - [2/3 -1/4] = -5/12 Therefore 5/6 - 5/12 = 5/12 units Finally the last on The square with vertices (1,1), (-1,1), (-1,-1), (1,-1) traversed once clockwise. To make this easier and doing 4 seperate intergrations, we can just multiply the Integration by four and since this is a nonconservative force it can't equal zero. so therefore the intergral from 1,1 to -1,1 times four is the amount of work done. For the last problem the line segment from (1,1) to (0,1) is -5/12. Multiply that by 2 and then that by 8, we have our answer, which is equal to -3.33 or -10/3 and the negative is just simply because we are going in a counterclockwise direction confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ********************************************* Question: Verify that the integral Int[(xy cos (xy) + sin (xy))dx + (x^2 cos(xy)) dy, C] is independent of path and evaluate it where C is any path from (0, pi/18) to (1, pi/6). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Int[(xy cos (xy) + sin (xy))dx + (x^2 cos(xy)) dy, C] M = (xy cos (xy) + sin (xy)) N = (x^2 cos(xy)) dM/dy = -x^2ysin(xy) + xcos(xy) + cos(xy) dN/dx = -x^2ysin(xy) + 2xcos(xy) Therefore the vector field is conservative Int[xycos(xy) + sin(xy)dx u=xy, dv = cos(xy)dx du = ydx, v = sin(xy)/y xsin(xy) - (2/y)cos(xy) + g(y) Int[ x^2cos(xy) = xsin(xy) + h(y) potential function: xsin(xy) Int[ xsin(xy) from (0, pi/18) to (1, pi/6) 1sin(1*pi/6) - 0 = 1/2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: This integral is of the form F dot ds for F = (x y cos(x y) + sin(x y) ) i + x^2 cos(x y) j, of the form M i + N j. We easily verify that curl F = (N_x - M_y) k = ( (2 x cos(xy) - x^2 y sin(x y)) - (x cos(x y) - x^2 y sin(x y) + x cos(x y) ) ) k = 0. We conclude that a scalar potential function exists, and we integrate to find it: Integrating x y cos(x y) + sin(x y) with respect to x we get cos( x y ) / y + x sin(x y) - cos(x y) / y = x sin(x y), to which we can add the arbitrary integration 'constant' g(y). Integrating x^2 cos(x y) with respect to y we get x^2 ( sin(x y) / x) = x sin(x y) + h(x). Our general scalar potential is therefore f(x, y) = x sin(x y) + g(y) + h(x), for arbitrary g(y) and h(x). The simplest scalar potential function is just f(x, y) = x sin(x y), and we will use this to evaluate our integral. For conservative field F our fundamental theorem says that integral ( F dot ds , between points (x0, y0) and (xf, yf) ) = f(xf, yf) - f(x0, y0) Our current integral is conservative. Its path originates at (0, pi/18) and terminates at (1, pi/6), so Int[(xy cos (xy) + sin (xy))dx + (x^2 cos(xy)) dy, C] = f(1, pi/6) - f(0, pi/18) = 1 sin(1 * pi / 6) - 0 sin(0 * pi/18) = 1/2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!#*&!
#$&* course Mth 277 1210pm 12/8/2011 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: The vector field is conservative if, and only if, its curl is zero. The curl of the field M i + N j + P k is (P_y - N_z) i + (M_z - P_x) j + (N_x - M_y) k. For the current function there is no z dependence and P = 0, so the expression becomes just (N_x - M_y) k. M_y = 1 and N_x = 2, so the field is not conservative. If we did try to find the scalar potential function, we would try to find f(x, y) such that del f = F. If this was so then we would have f_x = M and f_y = N. If f_x = (y - x^2), then integrating with respect to x we conclude that f = y x - x^3 / 3 + g(y), where g(y) is any function of y. If f_y = (2 x + y^2), then integrating with respect to y we conclude that f = 2 x y + y^3 / 3 + h(x), where h(x) is any function of x. We could let g(y) = y^3 / 3, and h(x) = - x^3 / 3. However the first integration yields y x, and the second yields 2 x y, both functions of x as well as y. That is, any f function such that del f = F must contain the term x y as well as the term 2 x y. We can't accomplish this by adjusting either g(y) or h(x), and we conclude that no scalar potential function f exists for this F. Had the original function been F(x, y) = (2 y - x^2) i + (2 x + y^2) j then : Our integrals would have been 2 y x - x^3 / 3 + g(y) and 2 x y + y^3 / 3 + h(x). Letting g(y) = y^3 / 3 and h(x) = -x^3 / 3, our function would be the same using either the x or the y integral; either way we would get 2 x y - x^3 / 3 + y^3 / 3. M_y would have been 2 and N_x would have been 2, so that our curl would have been (N_x - M_y) k = 0. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ********************************************* Question: Show that F = e^-y i - xe^-y j is conservative and find a scalar potential f for F. Evaluate the line integral Int[ F dot dR, C], where C is any smooth path connecting (0,0) and (1,1). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: M = e^-y N = -xe^-y dM/dy = -e^-y dN/dx = -e^-y Therefore the vector field is conservative f(x,y) = Int[f_x(x,y)dx] = Int[e^-y dx] = xe^-y + C1 f(x,y) = Int[f_y(x,y)dy] = Int[-xe^-y dy] = xe^-y + C2 f(x,y) = xe^-y Int[F dot dR, C] R = ti + tj, where 0 =< t =< 1 dR = i + j Int[(e^-y i - xe^-y j) dot (i + j)dt = Int[(e^-t - te^-t) dt from 0 to 1 u = -t, dv = e^-tdt du = dt, v = -e^-t = -e^-t + te^-t + Int[e^-tdt from 0 to 1 = -e^-t + te^-t - te^-t from 0 to 1 = -e^-t from 0 to 1 = -e^-1 + e^0 =appr 0.632 = -te^-t confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The curl of this function, which is of the form M i + N j + P k with M = e^-y, N = - x e^(-y) and P = 0, is easily found to be (P_y - N_z) i + (M_z - P_x) j + (N_x - M_y) k = (N_x - M_y) k = ( -e^-y - (-e^(-y)) ) k = 0. So the function is conservative and a scalar potential function exists. We therefore integrate: Integrating e^-y with respect to x we get x e^-y + g(y). Integrating - x e^(-y) with respect to y we get x e^-y + h(x). The general form of a scalar potential for our given vector field F is therefore f(x, y) = x e^-y + g(y) + h(x), where g(y) and h(x) are arbitrary functions of y and x, respectively. The simplest such function is obtained by letting g(y) and h(x) both be zero, giving us f(x, y) = x e^-y We can easily verify that del f = F. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ********************************************* Question: Consider the integral Int[ 2x^2y dx + x^3 dy, C]. Evaluate it on each of the following curves C: The semicircle x = sqrt(1-y) from -1 <= y <= 1. The line segment from (0,-1) to (1,1) and then the line segment from (1,1) to (0,1). The square with vertices (1,1), (-1,1), (-1,-1), (1,-1) traversed once clockwise. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First check to see F is conservative M = 2x^2*y N = x^3 My = 2x^2 Nx = 3x^2 NOT conservative, so now we have to do each on individually Int[(2x^2)*ydx + x^3dy y(t) = t from -1 to 1 x(t) = sqrt(1-t) dy = dt dx = [-1/sqrt(1-t)]dt Int[(2x^2)*ydx + x^3dy = Int{2t(1-t) - [1/sqrt(1-t)]dt = (2t^2)/2 - (2t^3)/3 + sqrt(1-t) + (2/5)(1-t)^(5/2) from 1 to -1 = [1 - 2/3 + 0 + 0] - [1 + 2/3 + sqrt(2) + (2/5)sqrt(32)] = -1 + sqrt(2) + (4/5)sqrt(8) =app -4.68 units, which is only negative because we are going from 1 to -1. Int[(2x^2)*ydx + x^3dy y - y1 = m(x - x1) y = 2x - 1 y(t) = 2t - 1 x(t) = t dy = 2dt dx = dt The line segment from (0,-1) to (1,1) Int[(2x^2)*ydx + x^3dy = 2t^2(2t-1)dt + 2t^3dt = 4t^3 - 2t^2dt + 2t^3dt =t^4 - (2/3)t^3 + (t^4)/2 from 1 to 0 = 1-(2/3)+(1/2) - 0 = 5/6 and then the line segment from (1,1) to (0,1) Int[(2x^2)*ydx + x^3dy y=1 x = -t + 1 dy = dt dx=-dt Int[(2x^2)*ydx + x^3dy = Int[(2(-t+1)^2)*-dt + (-t+1)^3dt from 0 to 1 =(2/3)(-t+1)^3 - (1/4)(-t+1)^4 from 0 to 1 = 0 - [2/3 -1/4] = -5/12 Therefore 5/6 - 5/12 = 5/12 units Finally the last on The square with vertices (1,1), (-1,1), (-1,-1), (1,-1) traversed once clockwise. To make this easier and doing 4 seperate intergrations, we can just multiply the Integration by four and since this is a nonconservative force it can't equal zero. so therefore the intergral from 1,1 to -1,1 times four is the amount of work done. For the last problem the line segment from (1,1) to (0,1) is -5/12. Multiply that by 2 and then that by 8, we have our answer, which is equal to -3.33 or -10/3 and the negative is just simply because we are going in a counterclockwise direction confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ********************************************* Question: Verify that the integral Int[(xy cos (xy) + sin (xy))dx + (x^2 cos(xy)) dy, C] is independent of path and evaluate it where C is any path from (0, pi/18) to (1, pi/6). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Int[(xy cos (xy) + sin (xy))dx + (x^2 cos(xy)) dy, C] M = (xy cos (xy) + sin (xy)) N = (x^2 cos(xy)) dM/dy = -x^2ysin(xy) + xcos(xy) + cos(xy) dN/dx = -x^2ysin(xy) + 2xcos(xy) Therefore the vector field is conservative Int[xycos(xy) + sin(xy)dx u=xy, dv = cos(xy)dx du = ydx, v = sin(xy)/y xsin(xy) - (2/y)cos(xy) + g(y) Int[ x^2cos(xy) = xsin(xy) + h(y) potential function: xsin(xy) Int[ xsin(xy) from (0, pi/18) to (1, pi/6) 1sin(1*pi/6) - 0 = 1/2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: This integral is of the form F dot ds for F = (x y cos(x y) + sin(x y) ) i + x^2 cos(x y) j, of the form M i + N j. We easily verify that curl F = (N_x - M_y) k = ( (2 x cos(xy) - x^2 y sin(x y)) - (x cos(x y) - x^2 y sin(x y) + x cos(x y) ) ) k = 0. We conclude that a scalar potential function exists, and we integrate to find it: Integrating x y cos(x y) + sin(x y) with respect to x we get cos( x y ) / y + x sin(x y) - cos(x y) / y = x sin(x y), to which we can add the arbitrary integration 'constant' g(y). Integrating x^2 cos(x y) with respect to y we get x^2 ( sin(x y) / x) = x sin(x y) + h(x). Our general scalar potential is therefore f(x, y) = x sin(x y) + g(y) + h(x), for arbitrary g(y) and h(x). The simplest scalar potential function is just f(x, y) = x sin(x y), and we will use this to evaluate our integral. For conservative field F our fundamental theorem says that integral ( F dot ds , between points (x0, y0) and (xf, yf) ) = f(xf, yf) - f(x0, y0) Our current integral is conservative. Its path originates at (0, pi/18) and terminates at (1, pi/6), so Int[(xy cos (xy) + sin (xy))dx + (x^2 cos(xy)) dy, C] = f(1, pi/6) - f(0, pi/18) = 1 sin(1 * pi / 6) - 0 sin(0 * pi/18) = 1/2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!#*&!#*&!